Past exam of the algebra course of the University of Oxford / 2009 / B

For the Abelian case of Cauchy theorem for groups, use induction on $|G|$. Choose $a\ne e$. If $p$ divides $m=|a|$, then $a^{m/p}$ has order $p$. Otherwise $⟨a⟩$ is a nontrivial normal subgroup of order prime to $p$, and $G/⟨a⟩$ has smaller order still divisible by $p$. Induction gives a coset of order $p$; any lift $b$ has order divisible by $p$, because the order of its image divides its own order. A suitable power of $b$ then has order $p$. This uses no classification of finite Abelian groups.

The map $g\mapsto g^{-1}$ partitions $S$ into disjoint two-element sets $\{g,g^{-1}\}$, so $|S|$ is even. Since $|G|$ is even, $|G\setminus S|$ is even as well. This complement contains the identity and consists of elements with $g^2=e$. It therefore contains at least one further element $h\ne e$, and that element has order two. Thus every finite group of even order contains an involution, without using Cauchy theorem for groups.

By Lagrange's theorem, a nonidentity element in a group of order four has order two or four. In the latter case the group is cyclic. Otherwise every nonidentity element has order two, and $(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$ while $(ab{)}^{-1}=ab$, so the group is Abelian. Choose distinct nonidentity elements $a,b$. Their cyclic subgroups meet trivially and $ab$ is the fourth element, so $(i,j)\mapsto a^i{b}^j$ gives an isomorphism $C_2\times C_2\to G$. HenceNo classification of finite Abelian groups is used.

By Cauchy theorem for groups, choose $a$ of order $p$ and $b$ of order two. The subgroup $P=⟨a⟩$ has index two, hence is normal, and $b\notin P$. Thus $a,b$ generate $G$. Conjugation by $b$ has $ba{b}^{-1}=a^r$ with $r\not\equiv 0(\mathrm{mod}p)$ and $r^2\equiv 1(\mathrm{mod}p)$. Since $p$ is prime, $r\equiv 1$ or $-1$.

If $r=1$, the two generators commute and $ab$ has order $2p$, so $G$ is cyclic. If $r=-1$, then $(ab{)}^2=e$ and $ab\ne b$. The two involutions $b,ab$ generate $G$, because $(ab)b=a$. Hence $G$ is dihedral in the stated convention. Thuswhere $D_{2p}$ denotes the dihedral group of order $2p$. In both cases $P$ is a proper nontrivial normal subgroup, so $G$ is not simple.

For a finite group with $|G|=p^{a}m$ and $p\nmid m$, a Sylow p-subgroup has order $p^a$. The Sylow theorems assert existence, containment of every p-subgroup in a Sylow subgroup, conjugacy of all Sylow subgroups, and $n_p\mid m$, $n_p\equiv 1(\mathrm{mod}p)$.

For Sylow subgroups of A6, $|A_6|=6!/2=360=2^3\cdot 3^2\cdot 5$. Explicit choices areTheir elements are even permutations, and their orders are the largest powers of five and three dividing $360$.

Every nonidentity element of $P_5$ is a five-cycle. There are $\left(\genfrac{}{}{0ex}{}{6}{5}\right)4!=144$ five-cycles, and distinct subgroups of order five meet only at the identity. Each contributes four such elements, givingFor $P_3$, each unordered partition of the six letters into two triples gives the subgroup generated by the two triple-cycles. It is independent of the chosen orientation of either cycle. There are $\frac{1}{2}\left(\genfrac{}{}{0ex}{}{6}{3}\right)=10$ such partitions. Different partitions give different subgroups, since the four single three-cycles in a subgroup identify the two triples. Every Sylow subgroup of order nine is conjugate to the displayed $P_3$, so every one arises from a partition in this way. HenceThe normalizer is $N_G(H)=\{g\in G:gH{g}^{-1}=H\}$. Its definition makes $H$ normal in $N_G(H)$. If $H$ is a Sylow subgroup of $G$, it is also a Sylow subgroup of $N_G(H)$, since $H\le N_G(H)\le G$ leaves no room for a larger power of $p$. The Sylow theorems make all Sylow subgroups of the normalizer conjugate there, and normality of $H$ makes every such conjugate equal to $H$. Thus $H$ is the unique Sylow subgroup of its normalizer.

The commutator subgroup is $G^{\prime }=⟨xy{x}^{-1}{y}^{-1}:x,y\in G⟩$. It is characteristic and hence normal, because every automorphism sends a commutator to a commutator. If $G^{\prime }\le H$, then $gh{g}^{-1}{h}^{-1}\in H$ for every $g\in G,h\in H$, so $H$ is normal. Every commutator becomes trivial in $G/H$, making that quotient Abelian. Conversely, the quotient homomorphism to an Abelian group kills each commutator, and hence kills $G^{\prime }$. ThereforeSet $G^{(0)}=G$, $G^{(i+1)}=(G^{(i)}{)}^{\prime }$. The group is soluble exactly when this derived series reaches the identity after finitely many steps.

If a nontrivial simple group is soluble, its normal subgroup $G^{\prime }$ is either $1$ or $G$. The latter would make the derived series constant and nontrivial, so $G^{\prime }=1$ and $G$ is Abelian. Every nonidentity element then generates $G$, since its cyclic subgroup is normal. An infinite cyclic group is not simple, and a finite cyclic group is simple exactly when its order is prime. This proves soluble simple group has prime order.

An Abelian group has the central series $1<G$, since $Z(G)=G$. In any central series, each factor is a subgroup of a center and hence Abelian. A finite normal series with Abelian factors makes the group soluble: inductively, if $N$ is soluble and $G/N$ is Abelian, then $G^{\prime }\subseteq N$, and successive derived groups eventually vanish. Hence every nilpotent group is soluble.

The group $S_3$ has the normal series $1<A_3<S_3$ with cyclic factors of orders three and two, so it is soluble. Its center is trivial, since a permutation commuting with both $(12)$ and $(23)$ must fix all three letters. Every nontrivial nilpotent group has nontrivial center: its first nonidentity central-series term is contained in $Z(G)$. Consequently $S_3$ is soluble but not nilpotent.

For a finite p-group, use induction on its order. The identity group is the base case. A nontrivial $G$ has nontrivial $Z=Z(G)$, so $G/Z$ has smaller order, still a power of $p$. By induction it has a central series. Lift that series to subgroups of $G$ and prepend $1<Z$. The first factor is central by definition of $Z$, and centrality of every lifted factor is exactly centrality of the corresponding quotient factor. This gives a central series for $G$. This proves finite p-groups are nilpotent: